ML Aggarwal Solution Class 10 Chapter 3 Shares and Dividends Test

Test

Question 1

If a man received ₹1080 as dividend from 9% ₹20 shares, find the number of shares purchased by him.

Sol :
Income on one share $=\frac{g}{10 n} \times 20$

$=\operatorname{Rs} \frac{9}{5}$

∴No. of shares $=1080 \times \frac{5}{9}$

= 120 x 5 = 600


Question 2

Find the percentage interest on capital invested in 18% shares when a Rs 10 share costs Rs 12.

Sol :

Dividend on one share = 18% of Rs 10

$=\frac{18 \times 10}{100}$
$=Rs \frac{9}{5}$

Income on $₹ 12=\frac{9}{5}$

Then income of $₹ 100=\frac{9}{5} \times \frac{100}{12}=15$

∴Percentage interest on capital=15 %


Question 3

Rohit Kulkami invests Rs 10000 in 10% Rs 100 shares of a company. If his annual dividend is Rs 800, find :

(i) The market value of each share.

(ii) The rate percent which he earns on his investment.

Sol :

Investment = Rs 10000

Face value of each share = Rs 100

Rate of dividend = 10%

Annual dividend = Rs 800

∴ Market value $=\frac{10000 \times 10}{800}=₹ 125$

(ii) Rate percent on investment

$=\frac{800 \times 100}{10000}=8 \%$


Question 4

At what price should a 9% Rs 100 share be quoted when the money is worth 6% ?

Sol :

If interest is 6 then investment = Rs 100

and if interest is 9, then investment

$=\operatorname{Rs} \frac{100 \times 9}{6}$
=Rs 150

Market value of each share = Rs 150


Question 5

By selling at Rs 92, some 2.5% Rs 100 shares and investing the proceeds in 5% Rs 100 shares at Rs 115, a person increased his annual income by Rs 90. Find:

(i) the number of shares sold.

(ii) the number of shares purchased.

(iii) the new income.

(iv) the rate percent which he earns on his investment.

Sol :

Rate of dividend = 2.5% and market price = Rs 92

Let number of shares purchased = x.

Selling price of x shares = 92 x

Income from investing

$₹ x=\frac{92 x \times 2 \cdot 5}{92}$
$=\frac{92 x \times 25}{92 \times 10}=\frac{5}{2} x$

Again by investing 92x in 5% at 115

The dividend $=\frac{92 x \times 5}{115}=4 x$

Difference $=4 x-\frac{5}{2} x=\frac{3}{2} x$

∴$\frac{3}{2} x=90$

⇒$x=\frac{90 \times 2}{3}=60$

(i) ∴No. of shares=60

(ii) No. of shares sold $=\frac{92 x}{115}$

$\frac{92 \times 60}{115}=48$

(iii) New income =4x=4×60=₹ 240

(iv) Rate percent interest on investment

$=\frac{5 \times 100}{115}=\frac{100}{23}$

$=4 \frac{8}{23} \%$


Question 6

A man has some shares of Rs. 100 par value paying 6% dividend. He sells half of these at a discount of 10% and invests the proceeds in 7% Rs. 50 shares at a premium of Rs. 10. This transaction decreases his income from dividends by Rs. 120. Calculate:

(i) the number of shares before the transaction.

(ii) the number of shares he sold.

(iii) his initial annual income from shares.

Sol :

Let no. of shares = x

Value of x shares = x × 100 = 100 x

and dividend$=\frac{100 x \times 6}{100}=\mathrm{Rs} .6 x$

and dividend on half-shares $=$ Rs. $\frac{6 x}{2}=$ Rs. $3 x$

Now, no. of shares he sold out $=\frac{x}{2}$

Amount received at $10 \%$ discount

$=\frac{x}{2} \times 90=$ Rs. $45 x$

In investing Rs 45x, no. shares he purchased$=\frac{45 x}{60}$

∴Amount of shares $=\frac{45 x}{60} \times 50=$ Rs. $\frac{225 x}{6}$

Income at the rate of $7 \%=\frac{225}{6} x \times \frac{7}{100}=\frac{21 x}{8}$

Difference in income$=3 x-\frac{21 x}{8}=\frac{3 x}{8}$

According to the condition, $\frac{3 x}{8}=120$

$x=\frac{120 \times 8}{3}=320$

(i) ∴No. of share he hold initially =320

(ii) No. of shares he hold later $=\frac{320}{2}=160$

(iii) Amount of income initially =320×6

=Rs 1920


Question 7

Divide Rs. 101520 into two parts such that if one part is invested in 8% Rs. 100 shares at 8% discount and the other in 9% Rs. 50 shares at 8% premium, the annual incomes are equal.

Sol :

Total investment = Rs. 101520

Let investment in first part = x

and in second part = (101520 – x)

Market value of first kind of shares = Rs. 100 – Rs. 8

= Rs. 92

and rate of dividend = 8%

∴Dividend $=\frac{x \times 8}{92}=\operatorname{Rs} . \frac{2 x}{23}$

Marked value of second kind=(101520-x)

Rate of dividend=9%

and Market value= Rs. $\frac{100+8}{100} \times 50=\frac{108}{100} \times 50$

=Rs 54

∴Dividend $=(101520-x) \times \frac{9}{2 \times 54}=\frac{101520-x}{12}$

∴According to the sum $\frac{2 x}{23}=\frac{101520-x}{12}$

⇒24x=101520×23-23x

⇒24x+23x=101520×23

⇒47x=101520×23

∴$x=\frac{101520 \times 23}{47}=49680$

∴Investment of first part=Rs 49680 and in second part

=Rs 1015620-Rs 49680=Rs 51840


Question 8

A man buys Rs. 40 shares of a company which pays 10% dividend. He buys the shares at such a price that his profit is 16% on his investment. At what price did he buy each share ?

Sol :

Face value of each share = Rs. 40

Dividend = 10%

Gain on investment = 10%

∴Dividend on Rs. $40=\frac{40 \times 10}{100}=\operatorname{Rs} .4$

Now Rs. 16 is interest on the market value=Rs 100

∴Market value if interest is Rs. 4

$=\frac{100 \times 4}{16}=$ Rs. 25

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more